Volume 59, 2013, 1–104
Robert Hakl, Alexander Lomtatidze, and Ioannis P. Stavroulakis
BOUNDED SOLUTIONS TO THE FIRST ORDER
SCALAR FUNCTIONAL DIFFERENTIAL EQUATIONS
Dedicated to the blessed memory of Professor Levan Magnaradze
ties of a bounded on [a,+∞[ solution to the boundary value problem for the first order functional differential equation is studied. The work is divided into two parts: in the first one, the linear problem is studied, the second one is devoted to the nonlinear problem.
2010 Mathematics Subject Classification. 34K10, 34K12, 34K06, 34K38.
Key words and phrases. Bounded solution, boundary value problem, functional-differential equation, solvability, uniqueness.
æØ . Œ ºØ ß ªŁ Ł [a,+∞[ Œ ª œ غø ØæŁ
ª Ł Ł æŁ æŒ ø ºŒ Łæ Æ Œø Łæ Œ ºŁ
غ ª æŁ ØºŒ Œ º , Æ º Æ Œ Œ-ØæÆØ ªº
. Œ ŁæŁ º º ø ß ª , ª ß ª Œ ºŁ .
Introduction
In the present work, the problem on the existence, uniqueness, and sign properties of a bounded on [a,+∞[ solution to the functional differential equation
u0(t) =F(u)(t) (0.1)
is studied.
In Chapter 1 (Sections 1–5) we consider a linear equation, i.e., the equation
u0(t) =`(u)(t) +q(t), (0.2)
with a “boundary” condition
ω(u) =c.
Here ` : Cloc([a,+∞[ ;R) → Lloc([a,+∞[ ;R) is a linear continuous op- erator, ω : Cloc([a,+∞[ ;R) → R is a linear bounded functional, c ∈ R, q∈Lloc([a,+∞[ ;R), and
sup
½¯¯
¯ Zt
a
q(s)ds
¯¯
¯: t≥a
¾
<+∞.
Main results of this chapter are contained in Section 3. In Subsection 3.1, theorems on “Fredholmity” of above mentioned problem are presented (The- orems 3.1 and 3.2). Optimal (unimprovable) sufficient conditions of exis- tence and uniqueness of a bounded solution to (0.2) satisfying one of the conditions
u(a) =c, u(+∞) =c, u(a)−u(+∞) =c
are presented in Subsection 3.2. Sign properties of those solutions are dis- cussed in Subsection 3.3.
In Section 4, the results of Section 3 are concretized for a special case of the equation (0.2), for the equation with deviating arguments
u0(t) = Xm
k=1
³
pk(t)u(τk(t))−gk(t)u(µk(t))
´ +q(t),
3
wherepk, gk ∈L([a,+∞[ ;R+) andτk, µk: [a,+∞[→[a,+∞[ are measur- able functions.
Chapter 2 is devoted to the nonlinear equation (0.1) with a boundary condition
ω(u) =h(u).
Here h : Cloc([a,+∞[ ;R) → R is (in general) a nonlinear functional. In Section 7 we establish so–called principle of existence of a bounded solu- tion, which is our main technical tool to investigate the nonlinear problem.
Optimal sufficient conditions for existence and uniqueness of a bounded so- lution to (0.1) are established in Section 8. In Section 10 we concretize results obtained in Section 8 for the equation with deviating arguments of the form
u0(t) = Xm
k=1
³
pk(t)u(τk(t))−gk(t)u(µk(t))
´ + +f¡
t, u(t), u(ν1(t)), . . . , u(νn(t))¢ , where pk, gk ∈L([a,+∞[ ;R+), τk, µk : [a,+∞[→ [a,+∞[ are measurab- le functions, and f : [a,+∞[×Rn+1 → R is a function satisfying local Carath´eodory conditions.
Notation
N is a set of all natural numbers.
Ris a set of all real numbers, R+= [0,+∞[ .
C([a, b];R) is a Banach space of continuous functionsu: [a, b]→R with a normkukC= max{|u(t)|: a≤t≤b}.
C([a, b];R+) ={u∈C([a, b];R) : u(t)≥0 fort∈[a, b]}.
Cloc([a,+∞[ ;D), withD ⊆R, is a set of continuous functionsu: [a,+∞[→ Dwith a topology of uniform convergence on every compact subin- terval of [a,+∞[ .
Ifu∈Cloc([a,+∞[ ;R), then kuk= sup©
|u(t)|: t≥aª .
C0([a,+∞[ ;R) is a set of functionsu∈Cloc([a,+∞[ ;R) (with a topology of uniform convergence on every compact subinterval of [a,+∞[ ), for each of which there exists a finite limit
u(+∞)def= lim
t→+∞u(t).
C([a, b];e D), where D ⊆ R, is a set of absolutely continuous functions u : [a, b]→ D.
Celoc([a,+∞[ ;D), where D ⊆ R, is a set of functions u : [a,+∞[→ D, absolutely continuous on every compact subinterval of [a,+∞[ . Ce0([a,+∞[ ;D) =Celoc([a,+∞[ ;D)∩C0([a,+∞[ ;R), whereD ⊆R.
L([a, b];R) is a Banach space of Lebesgue integrable functionsp: [a, b]→R with a norm
kpkL= Zb
a
|p(s)|ds.
L([a, b];R+) =©
p∈L([a, b];R) : p(t)≥0 for almost allt∈[a, b]ª .
5
L([a,+∞[ ;D), where D ⊆ R, is a set of Lebesgue integrable functions p: [a,+∞[→ D.
Lloc([a,+∞[ ;D), where D ⊆ R, is a set of locally Lebesgue integrable functionsp: [a,+∞[→ D with a topology of convergence in a mean on every compact subinterval of [a,+∞[ .
ch is a set of nontrivial linear bounded functionalsω:Cloc([a,+∞[ ;R)→R.
W0is a set of nontrivial linear bounded functionalsω:C0([a,+∞[ ;R)→R.
H is a set of continuous functionals h : Cloc([a,+∞[ ;R) → R with the following property: for everyr >0 there existsMr∈R+ such that
|h(v)| ≤Mr for kvk ≤r.
Lab is a set of linear bounded operators` : C([a, b];R) → L([a, b];R), for each of them there existsη∈L([a, b];R+) such that
|`(v)(t)| ≤η(t)kvkC for almost all t∈[a, b], v∈C([a, b];R).
Pab is a set of operators`∈ Lab transforming a setC([a, b];R+) into a set L([a, b];R+).
Mabis a set of measurable functions τ: [a, b]→[a, b].
Kab is a set of continuous operators F : C([a, b];R) → L([a, b];R) satis- fying Carath´eodory conditions, i.e., for everyr >0 there existsqr∈ L([a, b];R+) such that
|F(v)(t)| ≤qr(t) for almost all t∈[a, b], kvkC≤r.
Pe is a set of linear operators ` : Cloc([a,+∞[ ;R) → Lloc([a,+∞[ ;R) which are continuous, transform a setCloc([a,+∞[ ;R+) into a set Lloc([a,+∞[ ;R+), and such that`(1)∈L([a,+∞[ ;R+).
Leis a set of linear operators`:Cloc([a,+∞[ ;R)→Lloc([a,+∞[ ;R) which are continuous and for which there exists an operator`∈Pesuch that
|`(v)(t)| ≤`(|v|)(t) for almost all t∈[a,+∞[, v∈Cloc([a,+∞[ ;R).
Mis a set of locally measurable functionsτ: [a,+∞[→[a,+∞[ .
K is a set of continuous operatorsF :Cloc([a,+∞[ ;R)→Lloc([a,+∞[ ;R) satisfying local Carath´eodory conditions, i.e., for everyr > 0 there existsqr∈Lloc([a,+∞[ ;R+) such that
|F(v)(t)| ≤qr(t) for almost all t∈[a,+∞[, kvk ≤r.
Kloc
¡[a,+∞[×A;B¢
, whereA⊆Rn (n∈N),B ⊆R, is a set of functions f : [a,+∞[×A → B satisfying local Carath´eodory conditions, i.e., for all x ∈ A, the function f(·, x) : [a,+∞[→ B is measurable on
every compact subinterval of [a,+∞[ ,f(t,·) :A→Bis a continuous function for almost allt∈[a,+∞[ , and for every r >0 there exists qr∈Lloc([a,+∞[ ;R+) such that
|f(t, x)| ≤qr(t) for almost all t∈[a,+∞[, x∈A, kxk ≤r.
K¡
[a,+∞[×A;B¢
, whereA ⊆ Rn (n ∈N), B ⊆R, is a set of functions f : [a,+∞[×A→B satisfying Carath´eodory conditions, i.e., for all x ∈ A, the function f(·, x) : [a,+∞[→ B is measurable on every compact subinterval of [a,+∞[ , f(t,·) : A → B is a continuous function for almost allt∈[a,+∞[ , and for every r >0 there exists qr∈L([a,+∞[ ;R+) such that
|f(t, x)| ≤qr(t) for almost all t∈[a,+∞[, x∈A, kxk ≤r.
χab is a characteristic function of the interval [a, b], i.e., χab(t) =
(1 for t∈[a, b], 0 for t6∈[a, b].
θb :Cloc([a,+∞[ ;R)→C0([a,+∞[ ;R), whereb∈]a,+∞[ , is an operator defined by
θb(u)(t) =χab(t)u(t) + (1−χab(t))u(b) for t∈[a,+∞[. Ifq∈Lloc([a,+∞[ ;R) andb∈]a,+∞[ , then
e
qb(t) =χab(t)q(t) for almost all t∈[a,+∞[. If`∈Leandb∈]a,+∞[ , then
e`b(u)(t) =χab(t)`(θb(u))(t) for almost all t∈[a,+∞[. IfF ∈ Kand b∈]a,+∞[ , then
Feb(u)(t) =χab(t)F(θb(u))(t) for almost all t∈[a,+∞[. Ifω∈ch, resp. ω∈ W0, andb∈]a,+∞[ , then
e
ωb(u) =ω(θb(u)).
Ifh∈ Handb∈]a,+∞[ , then
ehb(u) =h(θb(u)).
We will say that ` ∈ Lab is a t0−Volterra operator, where t0 ∈ [a, b], if for arbitrarya1 ∈ [a, t0], b1 ∈ [t0, b], a1 6= b1, andv ∈ C([a, b];R), satisfying the equality
v(t) = 0 for t∈[a1, b1],
we have
`(v)(t) = 0 for almost all t∈[a1, b1].
We will say that`∈Leis ana−Volterra operator if for arbitraryb∈[a,+∞[
andv∈Cloc([a,+∞[ ;R), satisfying the equality v(t) = 0 for t∈[a, b], we have
`(v)(t) = 0 for almost all t∈[a, b].
We will say that ` ∈ Le is an anti–Volterra operator, if for arbitrary b ∈ [a,+∞[ andv∈Cloc([a,+∞[ ;R), satisfying the equality
v(t) = 0 for t∈[b,+∞[, we have
`(v)(t) = 0 for almost all t∈[b,+∞[. Ifx∈R, then
[x]+= 1
2(|x|+x), [x]−= 1
2(|x| −x).
The uniform convergence in [a,+∞[ is meant as a uniform convergence on every compact subinterval of [a,+∞[ .
The equalities and inequalities between measurable functions are under- stood almost everywhere in an appropriate interval.
Linear Problem
1. Statement of the Problem
In this chapter, we will consider the problem on the existence, unique- ness, and nonnegativity of a bounded solution to the equation
u0(t) =`(u)(t) +q(t) (1.1)
satisfying the condition
ω(u) =c. (1.2)
Here`∈L,e q∈Lloc([a,+∞[ ;R),ω∈ch, resp. ω∈ W0, and c∈R.
By a solution to (1.1) we understand a functionu∈ Celoc([a,+∞[ ;R) satisfying the equality (1.1) almost everywhere in [a,+∞[ . By a solution to the problem (1.1), (1.2) we understand a solution to (1.1) which belongs to the domain ofω and satisfies (1.2).
Along with the problem (1.1), (1.2) we will consider the corresponding homogeneous problem
u0(t) =`(u)(t), (1.10)
ω(u) = 0. (1.20)
Note that the particular cases of conditions (1.2), resp. (1.20), are the following conditions:
u(a) =c, (1.3)
u(+∞) =c, (1.4)
u(a)−u(+∞) =c, (1.5)
resp.
u(a) = 0, (1.30)
u(+∞) = 0, (1.40)
u(a)−u(+∞) = 0. (1.50)
Letb > abe arbitrary but fixed. Define the operatorsϕ:C([a, b];R)→ C0([a,+∞[ ;R) andψ:L([a,+∞[ ;R)→L([a, b];R) by
ϕ(v)(t)def= v(ν(t)) for t∈[a,+∞[, (1.6) ψ(u)(t)def= b−a
(b−t)2u(ν−1(t)) for t∈[a, b[, (1.7)
9
where
ν(t) =b− b−a
1 +t−a for t∈[a,+∞[, (1.8) andν−1is the inverse function to ν, i.e.,
ν−1(t) =a+b−a
b−t −1 for t∈[a, b[. Introduce the operatorb`:C([a, b];R)→L([a, b];R) by
`(v)(t)b def= ψ¡
`(ϕ(v))¢
(t) for t∈[a, b]. (1.9) On account of the assumption`∈Leit is clear that the operatorb`is well de- fined. Moreover, it can be easily verified that`bis a linear bounded operator.
Analogously, the functionalbω:C([a, b];R)→R, defined by b
ω(v)def= ω(ϕ(v)), (1.10)
is a linear bounded functional (in both cases, whenω∈ch andω∈ W0).
Now assume that
q∈L([a,+∞[ ;R), (1.11)
b
qdef= ψ(q), (1.12)
and consider the problem (on the interval [a, b])
v0(t) =`(v)(t) +b q(t),b ω(v) =b c (1.13) and the corresponding homogeneous problem
v0(t) =`(v)(t),b ω(v) = 0.b (1.130) By a direct calculation, it is easy to verify that if v is a solution∗ to the problem (1.13) then u def= ϕ(v) is a bounded solution to (1.1), (1.2), and vice versa, if uis a bounded solution to the problem (1.1), (1.2), then u∈Ce0([a,+∞[ ;R), and the functionv defined by
vdef= ϕ−1(u), v(b)def= u(+∞)
is a solution to (1.13). Therefore, the following proposition holds:
Proposition 1.1. Let (1.11) be fulfilled. Then the problem (1.1),(1.2) has a unique bounded solution u if and only if the problem (1.13) has a unique solution v. Moreover, u ∈ C0([a,+∞[ ;R), ϕ(v) ≡u, and v(b) = u(+∞).
A particular case of Proposition 1.1 is the following
Proposition 1.2. The only bounded solution of the problem (1.10), (1.20) is a trivial solution if and only if the problem (1.130) has only a trivial solution.
∗Solutions are understood in the sense of Carath´eodory, i.e., as absolutely continuous functions which satisfy the differential equality almost everywhere in [a, b].
Furthermore, the following assertion is well-known from the general theory of boundary value problems for functional differential equations (see, e.g., [1,2,10,15,19]).
Proposition 1.3. The problem(1.13)is uniquely solvable if and only if the corresponding homogeneous problem (1.130)has only a trivial solution.
Consequently, from Propositions 1.1–1.3, it immediately follows Proposition 1.4. Let (1.11) be fulfilled. Then the problem (1.1),(1.2) has a unique bounded solution if and only if the only bounded solution of the homogeneous problem (1.10),(1.20)is a trivial solution.
Therefore, the question on the existence and uniqueness of a bounded solution to (1.1), (1.2) is equivalent (under the assumption (1.11)) to the question on the unique solvability of the homogeneous boundary value prob- lem (1.130) (on a finite interval [a, b]).
Remark 1.1. It follows from the Riesz–Schauder theory that if the prob- lem (1.130) has a nontrivial solution, then for every c ∈ R there exists b
q∈L([a, b];R) such that the problem (1.13) has no solution.
Below we will study the problem (1.1), (1.2) under the less than (1.11) restricted condition, when
sup
½¯¯
¯ Zt
a
q(s)ds
¯¯
¯: t≥a
¾
<+∞ (1.14)
is fulfilled.
The chapter is organized as follows: Main results are presented in Sec- tion 3. First, the analogy of Proposition 1.4 is proved in Subsection 3.1 (see Theorems 3.1 and 3.2). In Subsection 3.2, sufficient conditions for the exis- tence and uniqueness of a bounded solution to the equation (1.1) satisfying one of the conditions (1.3), (1.4), or (1.5) are established. The question on sign constant solutions to the problems (1.1), (1.k) (k = 3,4,5) is dis- cussed in Subsection 3.3. In Section 4 we concretize results of Section 3 for particular cases of the equation (1.1) – for the equations with deviating arguments:
u0(t) = Xm
k=1
pk(t)u(τk(t)) +q(t), (1.15)
u0(t) =− Xm
k=1
gk(t)u(µk(t)) +q(t), (1.16) and
u0(t) = Xm
k=1
³
pk(t)u(τk(t))−gk(t)u(µk(t))
´
+q(t), (1.17)
wherepk, gk ∈L([a,+∞[ ;R+),τk, µk∈ M (k= 1, . . . , m). Last section of the chapter – Section 5 – is devoted to the examples verifying the optimality of obtained results.
Auxiliary propositions contained in Section 2 play a crucial role in prov- ing the main results. Namely Lemmas 2.1 and 2.2 (see p. 12 and p. 13) state that the unique bounded solution to (1.1), resp. (1.10), is a uniform limit of a suitable sequence{ub}b>aof solutions to the problem
u0(t) =`eb(u)(t) +qeb(t), (1.18) e
ωb(u) =c, (1.19)
resp.
u0(t) =`eb(u)(t), (1.180) e
ωb(u) = 0 (1.190)
(for the definition ofe`b,ωeb, andeqb see p. 7).
As it was mentioned above, we suppose that ` ∈Leand the condition (1.14) is fulfilled. If`(v)(t)def= p(t)v(t), i.e., if the equation (1.1) is of the form
u0(t) =p(t)u(t) +q(t), (1.20) the assumption`∈Lemeans that
p∈L([a,+∞[ ;R). (1.21)
The equation (1.20) is a suitable type of the equation (1.1) to verify that both conditions`∈Le(i.e., (1.21)) and (1.14) are essential for boundedness of its solutions.
2. Auxiliary Propositions
2.1. Lemma on Existence of a Bounded Solution.
Lemma 2.1. Let the condition (1.14) be satisfied and let there exist ρ0>0 andb0∈]a,+∞[, such that for everyb≥b0 the equation(1.18) has a solution ub satisfying the inequality
kubk ≤ρ0. (2.1)
Then the equation (1.1) has at least one bounded solution u. Moreover, there exists a sequence{ubn}+∞n=1⊂ {ub}b≥b0 such that
n→+∞lim ubn(t) =u(t) uniformly in [a,+∞[. (2.2) Proof. In view of (2.1) and the condition`∈L, from (1.18) withe u=ub we get
|ub(t)−ub(s)| ≤ρ0
Zt
s
`(1)(ξ)dξ+ Zt
s
|q(ξ)|dξ for a≤s≤t.
Hence, the set {ub}b≥b0 is uniformly bounded and equicontinuous on ev- ery compact subinterval of [a,+∞[ . Therefore, according to Arzel`a–Ascoli lemma, there exist a sequence {ubn}+∞n=1 ⊂ {ub}b≥b0 and a function u ∈ Cloc([a,+∞[ ;R) such that lim
n→+∞bn = +∞and (2.2) is fulfilled.
Obviously,
θbn(ubn)(t) =ubn(t) for t≥a, n∈N,
and therefore the integration of (1.18) (withu=ubn) fromatot yields ubn(t) =ubn(a) +
Zt
a
`(ubn)(ξ)dξ+ Zt
a
q(ξ)dξ for t∈[a, bn], n∈N.
Consequently, with respect to (2.2) and the assumption`∈L,e u(t) =u(a) +
Zt
a
`(u)(ξ)dξ+ Zt
a
q(ξ)dξ for t≥a,
i.e., u∈Celoc([a,+∞[ ;R) and it is a solution to the equation (1.1). More-
over, from (2.1) and (2.2) we havekuk ≤ρ0. ¤
2.2. Lemmas on A Priori Estimates.
Lemma 2.2. Let the only bounded solution to the problem(1.10),(1.20) be a trivial solution. Let, moreover, the condition (1.14) be fulfilled. Then there existb0∈]a,+∞[ andr0>0 such that for everyb≥b0, the problem (1.18),(1.19)has a unique solutionub, and this solution admits the estimate
kubk ≤r0(|c|+q∗), where
q∗= sup
½¯¯
¯ Zt
a
q(s)ds
¯¯
¯: t≥a
¾
. (2.3)
To prove this lemma we need some auxiliary propositions.
Proposition 2.1. Let the problem (1.130)have only a trivial solution.
Then there exists r0>0 such that for every bq∈L([a, b];R) andc∈R, the solution† v to the problem (1.13) admits the estimate
kvkC≤r0 µ
|c|+ sup
½¯¯
¯ Zt
a
b q(s)ds
¯¯
¯: t∈[a, b]
¾¶
. (2.4)
Proof. Let
R×L([a, b];R) =©
(c,qb) : c∈R, bq∈L([a, b];R)ª
†The existence and uniqueness of such a solution is guaranteed by Proposition 1.3.
be a linear space with the norm k(c,bq)kR×L=|c|+ sup
½¯¯
¯ Zt
a
b q(s)ds
¯¯
¯: t∈[a, b]
¾ ,
and let Ω be an operator, which assigns to every (c,bq) ∈ R×L([a, b];R) the solution v to the problem (1.13). According to Proposition 1.3, the operator Ω is defined correctly. Moreover, according to Theorem 1.4 in [15], Ω : R×L([a, b];R) → C([a, b];R) is a linear bounded operator (see also [10, Theorem 3.2]). Denote byr0the norm of Ω. Then, clearly, for any (c,qb)∈R×L([a, b];R), the inequality
kΩ(c,qb)kC≤r0k(c,qb)kR×L
holds. Consequently, the solutionv= Ω(c,bq) to the problem (1.13) admits
the estimate (2.4). ¤
From Proposition 2.1 it immediately follows
Proposition 2.2. Let b ∈ ]a,+∞[ be such that the problem (1.180), (1.190) has only a trivial solution. Then there exists rb > 0 such that for every q ∈ Lloc([a,+∞[ ;R) satisfying (1.14) and c ∈ R the problem (1.18),(1.19)has a unique solutionub, and this solution admits the estimate kubk ≤rb(|c|+q∗), (2.5) whereq∗ is defined by (2.3).
Proposition 2.3. Let the only bounded solution to the problem (1.10), (1.20) be a trivial solution. Then there exists b0 ∈ ]a,+∞[ such that for every b≥b0, the problem (1.180),(1.190)has only a trivial solution.
Proof. Assume on the contrary that there exists an increasing sequence {bk}+∞k=1, lim
k→+∞bk= +∞, such that, for everyk∈N, the problem
u0(t) =e`bk(u)(t), ωebk(u) = 0 (2.6) has a nontrivial solutionuk. Obviously,
uk≡θbk(uk) (2.7)
and, without loss of generality, we can assume that
kukk= 1 for k∈N. (2.8)
Furthermore, according to (2.6)–(2.8), and the assumption`∈L, we havee
|uk(t)−uk(s)| ≤ Zt
s
|u0k(ξ)|dξ= Zt
s
|`ebk(uk)(ξ)|dξ≤
≤ Zt
s
|`(uk)(ξ)|dξ≤ Zt
s
`(1)(ξ)dξ for a≤s≤t, k∈N.
Consequently, the sequence of functions{uk}+∞k=1 is uniformly bounded and equicontinuous on every compact subinterval of [a,+∞[ . According to Arzel`a–Ascoli lemma we can assume, without loss of generality, that there existsu0∈Cloc([a,+∞[ ;R) such that
k→+∞lim uk(t) =u0(t) uniformly in [a,+∞[. (2.9) Moreover, on account of (2.8) we have
ku0k ≤1. (2.10)
On the other hand, sinceuk (k∈N) are solutions to (2.6), we obtain uk(t) =uk(a) +
Zt
a
`ebk(uk)(s)ds for t≥a, k∈N, (2.11) e
ωbk(uk) = 0, k∈N. (2.12) From (2.11), in view of (2.9), we get
u0(t) =u0(a) + Zt
a
`(u0)(s)ds for t≥a. (2.13) Thusu0∈Celoc([a,+∞[ ;R) and
|u0(t)−u0(s)|=
¯¯
¯¯ Zt
s
`(u0)(ξ)dξ
¯¯
¯¯≤ ku0k Zt
s
`(1)(ξ)dξ for a≤s≤t.
The last inequality (together with (2.10)) and the fact that`∈Pe, ensures that there exists a finite limitu0(+∞). Consequently, from (2.12), in view of (2.9), we obtain
ω(u0) = 0. (2.14)
Now (2.13) and (2.14) imply thatu0 is a bounded solution to the problem (1.10), (1.20). Therefore,
u0≡0. (2.15)
Since`∈L, we can choosee b∗∈]a,+∞[ such that Z+∞
b∗
`(1)(s)ds≤ 1
3. (2.16)
According to (2.9) and (2.15) there existsk0∈N such that
|uk(t)| ≤ 1
3 for t∈[a, b∗], k≥k0. (2.17) On the other hand, from (2.11), in view of (2.8), we have
|uk(t)−uk(b∗)| ≤ Zt
b∗
|`ebk(uk)(s)|ds≤ Z+∞
b∗
`(1)(s)ds for t≥b∗, k∈N.
Hence, on account of (2.16) and (2.17), we obtain
|uk(t)| ≤ 2
3 for t≥a, k≥k0,
which contradicts (2.8). ¤
Proof of Lemma 2.2. Since the only bounded solution to the problem (1.10), (1.20) is a trivial solution, then according to Proposition 2.3 there exists b∗ ∈]a,+∞[ such that for everyb≥b∗ the problem (1.180), (1.190) has only a trivial solution. Consequently, according to Proposition 2.2 there existsrb>0 such that the solutionub to (1.18), (1.19) admits the estimate (2.5).
If|c|+q∗= 0 then the conclusion of the lemma is evident. Let, therefore,
|c|+q∗6= 0 and assume on the contrary that there exists a sequence{bn}+∞n=1 such thatbn≥b∗ forn∈N, lim
n→+∞bn= +∞, and
kubnk> n(|c|+q∗) for n∈N. (2.18) Put
vn(t) = ubn(t)
kubnk for t≥a, n∈N. (2.19) Then
kvnk= 1 for n∈N (2.20)
and, in view of (1.18), (1.19), and (2.19), we have v0n(t) =`ebn(vn)(t) +eqbn(t)
kubnk for t≥a, n∈N, (2.21) e
ωbn(vn) = c
kubnk for n∈N. (2.22) Obviously,θbn(vn)≡vn. From (2.21), on account of (2.18), (2.20), and the assumption`∈L, we obtaine
|vn(t)−vn(s)| ≤ Zt
s
|v0n(ξ)|dξ≤
≤ Zt
s
`(1)(ξ)dξ+ 1
|c|+q∗ Zt
s
|q(ξ)|dξ for a≤s≤t, n∈N. (2.23) From (2.20) and (2.23) it follows that the sequence of functions {vn}+∞n=1 is uniformly bounded and equicontinuous on every compact subinterval of [a,+∞[ . According to Arzel`a–Ascoli lemma we can assume, without loss of generality, that there existsv0∈Cloc([a,+∞[ ;R) such that
k→+∞lim vk(t) =v0(t) uniformly in [a,+∞[. (2.24) Moreover, on account of (2.20), we have
kv0k ≤1. (2.25)
On the other hand, the integration of (2.21) fromatotyields vn(t) =vn(a) +
Zt
a
`ebn(vn)(s)ds+ 1 kubnk
Zt
a
e
qbn(s)ds for n∈N.
Hence, in view of (2.18) and (2.24), we get v0(t) =v0(a) +
Zt
a
`(v0)(s)ds for t≥a. (2.26)
Thusv0∈Celoc([a,+∞[ ;R) and
|v0(t)−v0(s)|=
¯¯
¯¯ Zt
s
`(v0)(ξ)dξ
¯¯
¯¯≤ kv0k Zt
s
`(1)(ξ)dξ for a≤s≤t.
The last inequality, together with (2.25) and the fact that` ∈ P, ensurese that there exists a finite limit v0(+∞). Consequently, from (2.18), (2.22), and (2.24) we also obtain
ω(v0) = 0. (2.27)
Now (2.26) and (2.27) imply that v0 is a bounded solution to the problem (1.10), (1.20). Therefore,
v0≡0. (2.28)
Since`∈L, we can choosee a0∈]a,+∞[ such that Z+∞
a0
`(1)(s)ds≤ 1
5. (2.29)
According to (2.24) and (2.28) there existsn0∈N such that
|vn(t)| ≤1
5 for t∈[a, a0], n≥n0. (2.30) On the other hand, from (2.21), in view of (2.18) and (2.20), we have
|vn(t)−vn(a0)| ≤
≤ Zt
a0
|`ebn(vn)(s)|ds+ 2q∗ kubnk ≤
+∞Z
a0
`(1)(s)ds+2
n for t≥a0. Hence, on account of (2.29) and (2.30), we obtain
|vn(t)| ≤ 4
5 for t≥a, n≥max{n0,5}
which contradicts (2.20). ¤
2.3. Boundary Value Problems on Finite Interval. The following assertions are results from [9], formulated in a suitable for us form.
Lemma 2.3([9, Theorem 4.4, p. 83]). Let`b=`b0−`b1withb`0,`b1∈ Pab, and let there exist a functionbγ∈C([a, b]; ]0,e +∞[ ) such that
b
γ0(t)≥b`0(bγ)(t) +`b1(1)(t) for t∈[a, b], b
γ(b)−bγ(a)<3.
Then the problem
v0(t) =`(v)(t),b v(a) = 0 (2.31) has only a trivial solution.
Lemma 2.4([9, Theorem 4.2, p. 82]). Let`b=`b0−`b1withb`0,`b1∈ Pab, and let
Zb
a
`b0(1)(s)ds <1,
Zb
a
`b1(1)(s)ds <1 + 2 vu uu t1−
Zb
a
`b0(1)(s)ds .
Then the problem (2.31)has only a trivial solution.
Lemma 2.5([9, Theorem 4.10, p. 86]). Let`b=`b0−`b1withb`0,`b1∈ Pab, and let there exist a functionbγ∈C([a, b]; ]0,e +∞[ ) such that
−bγ0(t)≥`b1(bγ)(t) +b`0(1)(t) for t∈[a, b], b
γ(a)−bγ(b)<3.
Then the problem
v0(t) =b`(v)(t), v(b) = 0 (2.32) has only a trivial solution.
Lemma 2.6([9, Theorem 4.8, p. 86]). Let`b=`b0−`b1withb`0,`b1∈ Pab, and let
Zb
a
`b1(1)(s)ds <1,
Zb
a
`b0(1)(s)ds <1 + 2 vu uu t1−
Zb
a
`b1(1)(s)ds .
Then the problem (2.32)has only a trivial solution.
Lemma 2.7([9, Theorem 4.1, p. 80]). Let`b=`b0−`b1withb`0,`b1∈ Pab, and let either
Zb
a
`b0(1)(s)ds <1,
Rb a
b`0(1)(s)ds 1−Rb
a
b`0(1)(s)ds
<
Zb
a
b`1(1)(s)ds <2 + 2 vu uu t1−
Zb
a
b`0(1)(s)ds ,
or
Zb
a
`b1(1)(s)ds <1, Rb
a
b`1(1)(s)ds 1−
Rb a
b`1(1)(s)ds
<
Zb
a
b`0(1)(s)ds <2 + 2 vu uu t1−
Zb
a
b`1(1)(s)ds .
Then the problem
v0(t) =`(v)(t),b v(a)−v(b) = 0 has only a trivial solution.
Lemma 2.8 ([9, Theorem 2.1, p. 17]). Let `b∈ Pab and let there exist b
γ∈C([a, b]; ]0,e +∞[ )such that b
γ0(t)≥`(bbγ)(t) for t∈[a, b].
Then every functionv∈C([a, b];e R), satisfying
v0(t)≥`(v)(t)b for t∈[a, b], v(a)≥0, (2.33) is nonnegative.
Lemma 2.9([9, Theorem 2.5, p. 22]). Let`b=`b0−`b1withb`0,`b1∈ Pab, and let`b1 be an a−Volterra operator. Let, moreover,
Zb
a
`b0(1)(s)ds <1, Zb
a
b`1(1)(s)ds≤1.
Then every functionv∈C([a, b];e R)satisfying (2.33) is nonnegative.
Remark 2.1. Note that under the conditions of Lemma 2.8 or Lem- ma 2.9, the problem
v0(t) =`(v)(t) +b q(t),b v(a) =c
is uniquely solvable for any qb ∈ L([a, b];R) and c ∈ R. Moreover, the solution is nonnegative wheneverqb∈L([a, b];R+) andc∈R+.
Indeed, let the assumptions of Lemma 2.8 or Lemma 2.9 be fulfilled and letv be a solution to (2.31). Then, obviously, −v is also a solution to (2.31), and so bothv and−v are nonnegative functions. Therefore,v≡0.
Thus the assertion follows from Proposition 1.3 and Lemmas 2.8 and 2.9.
Lemma 2.10 ([9, Theorem 2.12, p. 26]). Let −`b∈ Pab and let there exist bγ∈C([a, b]; ]0,e +∞[ ) such that
b
γ0(t)≤`(bbγ)(t) for t∈[a, b].
Then every functionv∈C([a, b];e R)satisfying
v0(t)≤b`(v)(t) for t∈[a, b], v(b)≥0 (2.34) is nonnegative.
Lemma 2.11([9, Theorem 2.16, p. 28]). Let `b=`b0−`b1 with`b0,b`1∈ Pab, and letb`0 be ab−Volterra operator. Let, moreover,
Zb
a
`b0(1)(s)ds≤1, Zb
a
b`1(1)(s)ds <1.
Then every functionv∈C([a, b];e R)satisfying (2.34) is nonnegative.
Remark 2.2. Analogously to Remark 2.1 it can be shown that under the conditions of Lemma 2.10 or Lemma 2.11, the problem
v0(t) =`(v)(t) +b bq(t), v(b) =c
is uniquely solvable for any qb ∈ L([a, b];R) and c ∈ R. Moreover, the solution is nonnegative whenever−bq∈L([a, b];R+) andc∈R+.
Lemma 2.12([9, Theorem 2.4, p. 21]).Letb`=`b0−`b1withb`0,`b1∈ Pab, and let
Zb
a
`b0(1)(s)ds <1, Rb a
b`0(1)(s)ds 1−
Rb a
b`0(1)(s)ds
<
Zb
a
b`1(1)(s)ds≤1.
Then every functionv∈C([a, b];e R)satisfying
v0(t)≥`(v)(t)b for t∈[a, b], v(a)−v(b)≥0 (2.35) is nonnegative.
Lemma 2.13([9, Theorem 2.15, p. 28]). Let `b=`b0−`b1 with`b0,b`1∈ Pab, and let
Zb
a
`b1(1)(s)ds <1, Rb a
b`1(1)(s)ds 1−Rb
a
b`1(1)(s)ds
<
Zb
a
b`0(1)(s)ds≤1.
Then every functionv∈C([a, b];e R)satisfying (2.35) is nonpositive.
Remark 2.3. Analogously to Remark 2.1 it can be shown that under the conditions of Lemma 2.12 or Lemma 2.13, the problem
v0(t) =`(v)(t) +b q(t),b v(a)−v(b) =c
is uniquely solvable for any qb ∈ L([a, b];R) and c ∈ R. Moreover, the solution is nonnegative, resp. nonpositive, whenever qb∈L([a, b];R+) and c∈R+.
2.4. Nonnegative Solutions to a Certain Differential Inequali- ty.
Lemma 2.14. Let`∈Peand let there existγ∈Celoc([a,+∞[ ; ]0,+∞[ ) such that
γ0(t)≥`(γ)(t) for t≥a, (2.36)
t→+∞lim γ(t) = +∞. (2.37)
Then every bounded functionu∈Celoc([a,+∞[ ;R), satisfying
u0(t)≥`(u)(t) for t≥a, u(a)≥0, (2.38) is nonnegative.
Proof. Assume on the contrary that there exist a bounded function u ∈ Celoc([a,+∞[ ;R) andt0> asuch thatusatisfies (2.38) and
u(t0)<0. (2.39)
Put
λ= sup
½
−u(t)
γ(t) : t≥a
¾
. (2.40)
Obviously, by virtue of (2.39),
λ >0. (2.41)
Moreover, in view of (2.37) and the assumption thatuis a bounded function, there existst1≥asuch that
λ=−u(t1)
γ(t1). (2.42)
Further, put
w(t) =λγ(t) +u(t) for t≥a.
Then, on account of (2.36), (2.38), (2.40)–(2.42), we have
w0(t)≥`(w)(t) for t≥a, (2.43) w(t)≥0 for t≥a, w(a)>0, (2.44)
w(t1) = 0. (2.45)
Now, due to (2.44) and the assumption`∈Pe, from (2.43) we getw0(t)≥0 fort≥a. Consequently,w(t)>0 fort≥a, which contradicts (2.45). ¤ Remark 2.4. Under the conditions of Lemma 2.14, the only bounded solution to the problem (1.10), (1.30) is a trivial solution. Indeed, let ube a bounded solution to (1.10), (1.30). Then, obviously,−uis also a bounded solution to (1.10), (1.30), and both u and −u are nonnegative functions.
Therefore,u≡0.
3. Main Results 3.1. Necessary and Sufficient Conditions.
Theorem 3.1. Let ω ∈chand the condition (1.14) be fulfilled. Then the problem(1.1),(1.2)has a unique bounded solution if and only if the only bounded solution to the problem (1.10),(1.20) is a trivial solution.
Proof. Evidently, if the problem (1.1), (1.2) has a unique bounded solution for arbitraryc∈Randq∈Lloc([a,+∞[ ;R) satisfying (1.14), then the only bounded solution to the problem (1.10), (1.20) is a trivial solution.
If the only bounded solution to the problem (1.10), (1.20) is a trivial solution, then, according to Lemmas 2.1 and 2.2, the equation (1.1) has at least one bounded solution u and there exists a sequence of functions {ubn}+∞n=1⊂Ce0([a,+∞[ ;R) such thatω(ubn) =c (n∈N), and (2.2) holds.
Consequently, since ω ∈ ch, we also have ω(u) = c, i.e., u is a bounded solution to the problem (1.1), (1.2). The uniqueness ofuis evident. ¤ Theorem 3.2. Let ω∈ W0 and the condition (1.14) be satisfied. Let, furthermore, the only bounded solution to the problem (1.10),(1.20) be a trivial solution. Then the equation (1.1) has at least one bounded solution.
If, moreover, there exists a finite limit
t→+∞lim Zt
a
q(s)ds, (3.1)
then the problem (1.1),(1.2)has a unique bounded solution.
Proof. If the assumptions of theorem are fulfilled, then, according to Lem- mas 2.1 and 2.2, the equation (1.1) has at least one bounded solutionu. Fur- thermore, there exist a sequence of functions {ubn}+∞n=1 ⊂ Ce0([a,+∞[ ;R) such that ω(ubn) = c (n ∈ N), and (2.2) holds. Thus, if there exists a finite limit (3.1), there also exists a finite limit u(+∞). Consequently,
u∈ C0([a,+∞[ ;R) and since ω ∈ W0, we also have ω(u) =c, i.e., u is a bounded solution to the problem (1.1), (1.2). In this case, the uniqueness
ofuis evident. ¤
3.2. Bounded Solutions. In the first part of this subsection, there are formulated theorems dealing with the existence and uniqueness of a bounded solution to the problems (1.1), (1.k) (k = 3,4,5). The proofs of those theorems can be found in the second part of this subsection.
Theorem 3.3. Let`=`0−`1,`0, `1∈Pe, and let the condition (1.14) be satisfied. Let, moreover, there existγ∈Ce0([a,+∞[ ; ]0,+∞[ ) such that
γ0(t)≥`0(γ)(t) +`1(1)(t) for t≥a, (3.2)
γ(+∞)−γ(a)<3. (3.3)
Then the problem (1.1),(1.3)has a unique bounded solution.
Remark 3.1. Theorem 3.3 is unimprovable in the sense that the strict inequality (3.3) cannot be replaced by the nonstrict one (see Example 5.1, p. 40).
Theorem 3.4. Let`=`0−`1,`0, `1∈Pe, and let the condition (1.14) be satisfied. Let, moreover,
+∞Z
a
`0(1)(s)ds <1, (3.4)
Z+∞
a
`1(1)(s)ds <1 + 2 vu uu t1−
+∞Z
a
`0(1)(s)ds . (3.5) Then the problem (1.1),(1.3)has a unique bounded solution.
Remark 3.2. Denote byGa the set of pairs (x, y)∈R+×R+such that x <1, y <1 + 2√
1−x (see Figure 3.1).
Theorem 3.4 states that if the condition (1.14) is fulfilled,`=`0−`1, where`0, `1∈Pe, and
µZ+∞
a
`0(1)(s)ds, Z+∞
a
`1(1)(s)ds
¶
∈Ga, then the problem (1.1), (1.3) has a unique bounded solution.
Below we will show (see On Remark 3.2, p. 40) that for every x0, y0∈ R+, (x0, y0)6∈ Ga there exist `0, `1 ∈Pe, q∈ Lloc([a,+∞[ ;R), and c ∈ R
x y
1 3
G
ay= 1 + 2√ 1−x
Figure 3.1
such that (1.14) holds,
+∞Z
a
`0(1)(s)ds=x0,
+∞Z
a
`1(1)(s)ds=y0,
and the problem (1.1), (1.3) with`=`0−`1has no solution. In particular, the strict inequalities (3.4) and (3.5) cannot be replaced by the nonstrict ones.
Theorem 3.5. Let`=`0−`1,`0, `1∈Pe, and let the condition (1.14) be satisfied. Let, moreover, there existγ∈Ce0([a,+∞[ ; ]0,+∞[ ) such that
−γ0(t)≥`1(γ)(t) +`0(1)(t) for t≥a, (3.6)
γ(+∞)>0, (3.7)
γ(a)−γ(+∞)<3. (3.8)
Then the equation (1.1) has at least one bounded solution. If, moreover, there exists a finite limit (3.1), then the problem (1.1),(1.4) has a unique solution.
Remark 3.3. Theorem 3.5 is unimprovable in the sense that neither one of the strict inequalities (3.7) and (3.8) can be replaced by the nonstrict one (see Example 5.2 on p. 41 and Example 5.3 on p. 42).
Theorem 3.6. Let`=`0−`1,`0, `1∈Pe, and let the condition (1.14) be satisfied. Let, moreover,
Z+∞
a
`1(1)(s)ds <1, (3.9)
Z+∞
a
`0(1)(s)ds <1 + 2 vu uu t1−
+∞Z
a
`1(1)(s)ds . (3.10) Then the equation (1.1) has at least one bounded solution. If, moreover, there exists a finite limit (3.1), then the problem (1.1),(1.4) has a unique solution.
Remark 3.4. Denote by G+∞ the set of pairs (x, y) ∈R+×R+ such that
y <1, x <1 + 2p 1−y (see Figure 3.2).
x y
3 1
G
+∞x= 1 + 2√1−y
Figure 3.2
Theorem 3.6 states that if there exists a finite limit (3.1), `=`0−`1, where`0, `1∈Pe, and
µ +∞Z
a
`0(1)(s)ds,
+∞Z
a
`1(1)(s)ds
¶
∈G+∞, then the problem (1.1), (1.4) has a unique bounded solution.
Below we will show (see On Remark 3.4, p. 43) that for every x0, y0∈ R+, (x0, y0)6∈G+∞ there exist`0, `1∈Pe,q∈Lloc([a,+∞[ ;R), and c∈R such that there exists a finite limit (3.1),
+∞Z
a
`0(1)(s)ds=x0,
+∞Z
a
`1(1)(s)ds=y0,
and the problem (1.1), (1.4) with`=`0−`1has no solution. In particular, the strict inequalities (3.9) and (3.10) cannot be replaced by the nonstrict ones.
Theorem 3.7. Let`=`0−`1,`0, `1∈Pe, and let the condition (1.14) be satisfied. Let, moreover, either
+∞Z
a
`0(1)(s)ds <1, (3.11)
+∞R
a
`0(1)(s)ds 1−+∞R
a
`0(1)(s)ds
<
+∞Z
a
`1(1)(s)ds <2 + 2 vu uu t1−
+∞Z
a
`0(1)(s)ds , (3.12)
or
+∞Z
a
`1(1)(s)ds <1, (3.13)
+∞R
a
`1(1)(s)ds 1−+∞R
a
`1(1)(s)ds
<
+∞Z
a
`0(1)(s)ds <2 + 2 vu uu t1−
+∞Z
a
`1(1)(s)ds . (3.14)
Then the equation (1.1) has at least one bounded solution. If, moreover, there exists a finite limit (3.1), then the problem (1.1),(1.5) has a unique solution.
Remark 3.5. Put G+p =
n
(x, y)∈R+×R+: x <1, x
1−x < y <2 + 2√ 1−x
o , G−p =
n
(x, y)∈R+×R+: y <1, y
1−y < x <2 + 2p 1−y
o
(see Figure 3.3).
Theorem 3.7 states that if there exists a finite limit (3.1), `=`0−`1, where`0, `1∈Pe, and
µ +∞Z
a
`0(1)(s)ds,
+∞Z
a
`1(1)(s)ds
¶
∈G+p ∪G−p, then the problem (1.1), (1.5) has a unique bounded solution.
Below we will show (see On Remark 3.5, p. 44) that for every x0, y0∈ R+, (x0, y0) 6∈G+p ∪G−p there exist`0, `1 ∈ P,e q∈ Lloc([a,+∞[ ;R), and c∈Rsuch that there exists a finite limit (3.1),
+∞Z
a
`0(1)(s)ds=x0,
+∞Z
a
`1(1)(s)ds=y0,
x y
4 4
y= 2 + 2√1−x
y=1−xx
x= 2 + 2√1−y x=1y
−y
G+p
G−p
Figure 3.3
and the problem (1.1), (1.5) with`=`0−`1has no solution. In particular, neither of the strict inequalities in (3.11)–(3.14) can be replaced by the nonstrict one.
Remark 3.6. In Theorems 3.1–3.7, the condition (1.14) is essential and it cannot be omitted. Indeed, letq∈Lloc([a,+∞[ ;R),p∈L([a,+∞[ ;R+) be such that
06=
+∞Z
a
p(s)ds <1, and consider the equation
u0(t) =p(t)u(a) +q(t). (3.15) Put
`(v)(t)def= p(t)v(a) for t≥a.
Then the assumptions imposed on the operator` (with`0≡` and`1 ≡0) in Theorems 3.3–3.7 are fulfilled.
On the other hand, every solutionuto (3.15) is of the form u(t) =c0
µ 1 +
Zt
a
p(s)ds
¶ +
Zt
a
q(s)ds for t≥a,
wherec0∈R. However,uis bounded if and only if the condition (1.14) is fulfilled.
Remark 3.7. It is clear that if the problem (1.1), (1.4), resp. (1.1), (1.5), has a solution for some c∈ R, then there exists a finite limit (3.1). Thus the condition (3.1) in Theorems 3.5–3.7 is also a necessary condition for the unique solvability of the mentioned problems.
